Effective mass (spring–mass system)

In a real spring–mass system, the spring has a non-negligible mass $m$. Since not all of the spring's length moves at the same velocity $u$ as the suspended mass $M$, its kinetic energy is not equal to $m u^2 /2$. As such, $m$ cannot be simply added to $M$ to determine the frequency of oscillation, and the effective mass of the spring is defined as the mass that needs to be added to $M$ to correctly predict the behavior of the system.

Ideal uniform spring

The effective mass of the spring in a spring-mass system when using an ideal spring of uniform linear density is 1/3 of the mass of the spring and is independent of the direction of the spring-mass system (i.e., horizontal, vertical, and oblique systems all have the same effective mass). This is because external acceleration does not affect the period of motion around the equilibrium point.

We can find the effective mass of the spring by finding its kinetic energy. This requires adding all the length elements' kinetic energy, and requires the following integral:

$T =\int_m\frac{1}{2}u^2\,dm$

Since the spring is uniform, $dm=\left(\frac{dy}{L}\right)m$, where $L$ is the length of the spring. Hence,

$T = \int_0^L\frac{1}{2}u^2\left(\frac{dy}{L}\right)m\!$

$=\frac{1}{2}\frac{m}{L}\int_0^L u^2\,dy$

The velocity of each mass element of the spring is directly proportional to its length, i.e. $u=\frac{vy}{L}$, from which it follows:

$T =\frac{1}{2}\frac{m}{L}\int_0^L\left(\frac{vy}{L}\right)^2\,dy$

$=\frac{1}{2}\frac{m}{L^3}v^2\int_0^L y^2\,dy$

$=\frac{1}{2}\frac{m}{L^3}v^2\left[\frac{y^3}{3}\right]_0^L$

$=\frac{1}{2}\frac{m}{3}v^2$

Comparing to the expected original kinetic energy formula $\frac{1}{2}mv^2,$ we can conclude that effective mass of spring in this case is m/3. Using this result, we can write down the equation of motion of the spring by writing the Lagrangian in terms of the displacement $x$ from the spring's unstretched position (ignoring constant potential terms and taking the upwards direction as positive):

$L = T - V$

$= \frac{1}{2}\frac{m}{3}\dot{x}^2 + \frac{1}{2}M \dot{x}^2 - \frac{1}{2} k x^2 - \frac{m g x}{2} - M g x$

Note that $g$ here is the acceleration of gravity along the spring. Using the Euler–Lagrange equation, we arrive at the equation of motion:

$\left( \frac{m}{3}+M \right) \ddot{x} = -kx - \frac{mg}{2} - Mg$

We can find the equilibrium point $x_{\mathrm{eq}}$ by letting the acceleration be zero:

$x_{\mathrm{eq}} = -\frac{1}{k}\left( \frac{mg}{2} + Mg \right)$

Defining $\bar{x} = x - x_{\mathrm{eq}}$, the equation of motion becomes:

$\left( \frac{m}{3}+M \right) \ddot{\bar{x}} = -k\bar{x}$

This is the equation for a simple harmonic oscillator with period:

$\tau = 2 \pi \left( \frac{M + m/3}{k} \right)^{1/2}$

So we can see that the effective mass of the spring added to the mass of the load gives us the "effective total mass" of the system that must be used in the standard formula $2 \pi \left( \frac{m}{k} \right)^{1/2}$ to determine the period of oscillation.

General case

As seen above, the effective mass of a spring does not depend upon "external" factors such as the acceleration of gravity along it. In fact, for a non-uniform spring, the effective mass solely depends on its linear density $\rho(x)$ along its length:

$T = \int_m\frac{1}{2}u^2\,dm$

$= \int_{0}^L\frac{1}{2}u^2 \rho(x) \,dx$

$= \int_{0}^L\frac{1}{2}\left(\frac{v x}{L} \right)^2 \rho(x) \,dx$

$= \frac{1}{2} \left[ \int_{0}^L \frac{x^2}{L^2} \rho(x) \,dx \right] v^2$

So the effective mass of a spring is:

$m_{\mathrm{eff}} = \int_{0}^L \frac{x^2}{L^2} \rho(x) \,dx$

This result also shows that $m_{\mathrm{eff}} \le m$, with $m_{\mathrm{eff}} = m$ occurring in the case of an unphysical spring whose mass is located purely at the end farthest from the support.

Real spring

The above calculations assume that the stiffness coefficient of the spring does not depend on its length. However, this is not the case for real springs. For small values of $M/m$, the displacement is not so large as to cause elastic deformation. Jun-ichi Ueda and Yoshiro Sadamoto have found^[1] that as $M/m$ increases beyond 7, the effective mass of a spring in a vertical spring-mass system becomes smaller than Rayleigh's value $m/3$ and eventually reaches negative values. This unexpected behavior of the effective mass can be explained in terms of the elastic after-effect (which is the spring's not returning to its original length after the load is removed).

See also

• Simple harmonic motion (SHM) examples.
• Reduced mass

References

1. http://journals.jps.jp/doi/abs/10.1143/JPSJ.66.367

External links

• http://tw.knowledge.yahoo.com/question/question?qid=1405121418180
• http://tw.knowledge.yahoo.com/question/question?qid=1509031308350
• http://hk.knowledge.yahoo.com/question/article?qid=6908120700201
• http://www.goiit.com/posts/list/mechanics-effective-mass-of-spring-40942.htm
• http://www.juen.ac.jp/scien/sadamoto_base/spring.html
• "The Effective Mass of an Oscillating Spring" Amer. J. Phys., 38, 98 (1970)
• "Effective Mass of an Oscillating Spring" The Physics Teacher, 45, 100 (2007)